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a: \(=y^2-9\)
b: \(=m^3+n^3\)
c: \(=8-a^3\)
d: \(=\left(a-b-c-a+b-c\right)\left(a-b-c+a-b+c\right)\)
\(=-2c\cdot\left(2a-2b\right)\)
\(=-4ac+4bc\)
f: \(=\left(1-x^3\right)\left(1+x^3\right)=1-x^6\)
b: \(\left(m+n\right)\times\left(m^2-mn+n^2\right)=m^3+n^3\)
\(a,\left(2a+3\right)x-\left(2a+3\right)y+\left(2a+3\right)\)
\(=\left(2a+3\right)\left(x-y+1\right)\)
\(b,\left(4x-y\right)\left(a-1\right)-\left(y-4x\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)
\(=\left(4x-y\right)\left(a-1\right)+\left(4x-y\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)
\(=\left(4x-y\right)\left(a-1+b-1+1-c\right)\)
\(=\left(4x-y\right)\left(a+b-c-1\right)\)
\(c,x^k+1-x^k-1\)
\(=0?!?!\)
\(d,x^m+3-x^m+1\)
\(=4\)
\(e,3\left(x-y\right)^3-2\left(x-y\right)^2\)
\(=\left(x-y\right)^2\left(3\left(x-y\right)-2\right)\)
\(=\left(x-y\right)^2\left(3x-3y-2\right)\)
\(f,81a^2+18a+1\)
\(=\left(9a\right)^2+2.9a+1\)
\(=\left(9a+1\right)^2\)
\(g,25a^2.b^2-16c^2\)
\(=\left(5ab\right)^2-\left(4c\right)^2\)
\(=\left(5ab+4c\right)\left(5ab-4c\right)\)
\(h,\left(a-b\right)^2-2\left(a-b\right)c+c^2\)
\(=\left(a-b-c\right)^2\)
\(i,\left(ax+by\right)^2-\left(ax-by\right)^2\)
\(=\left(ax+by-ax+by\right)\left(ax+by+ax-by\right)\)
\(=2by.2ax\)
\(=4axby\)
a) \(\dfrac{3x^2y}{2xy^5}=\dfrac{3x}{2y^4}\)
b) \(\dfrac{3x^2-3x}{x-1}=\dfrac{3x\left(x-1\right)}{x-1}=3x\)
c) \(\dfrac{ab^2-a^2b}{2a^2+a}=\dfrac{ab\left(b-a\right)}{a\left(2a+1\right)}=\dfrac{b\left(b-a\right)}{2a+1}=\dfrac{b^2-ab}{2a+1}\)
d) \(\dfrac{12\left(x^4-1\right)}{18\left(x^2-1\right)}=\dfrac{2\left(x^2-1\right)\left(x^2+1\right)}{3\left(x^2-1\right)}=\dfrac{2\left(x^2+1\right)}{3}\)
`a, (3x^2y)/(2xy^5)`
`= (3x)/(2y^4)`
`b, (3x^2-3x)/(x-1)`
`= (3x(x-1))/(x-1)`
`= 3x`
`c, (ab^2-a^2b)/(2a^2+a)`
`= (b(a-b))/((2a+1))`
`d, (12(x^4-1))/(18(x^2-1)) = (2(x^2+1))/3`.
Bài 12:
1) A = x2 - 6x + 11
= (x2 - 6x + 9) + 2
= (x - 3)2 + 2
Ta có: (x - 3)2 ≥ 0 ∀ x
Dấu ''='' xảy ra khi x - 3 = 0 ⇔ x = 3
Do đó: (x - 3)2 + 2 ≥ 2
Hay A ≥ 2
Dấu ''='' xảy ra khi x = 3
Vậy Min A = 2 tại x = 3
2) B = x2 - 20x + 101
= (x2 - 20x + 100) + 1
= (x - 10)2 + 1
Ta có: (x - 10)2 ≥ 0 ∀ x
Dấu ''='' xảy ra khi x - 10 = 0 ⇔ x = 10
Do đó: (x - 10)2 + 1 ≥ 1
Hay B ≥ 1
Dấu ''='' xảy ra khi x = 10
Vậy Min B = 1 tại x = 10
a, <=>y2-32 <=> y2 -9 (hằng đẳng thức số 3)
b, <=> m3+n3 ( hằng đẳng thức số 6)
c, <=> 23-a3 (__________________số 7)
d, <=> (a-b-c-a+b-c )( a-b-c+a-b+c)
<=> -2c*2a= -4ac
e, <=> (a-x-y-a-x+y) [(a-x-y) 2+(a-x-y)(a+x-y)+(a+x-y)2]
(Nhân phá ngoặc) -)
d <=> (1-x2)[(1+x2)2-x2)
<=> (1-x2)(1+2x2)
<=> 1+2x2-x2-2x4
<=> 1+x2-2x4